2015年11月2日 · # sin a ± sin b = 2 sin ((a±b)/2) cos ((a∓b)/2) # # tan A = sin A/cos A # # (sin 80^@ - sin 10^@)/(sin 80^@ + sin 10^@) = sin 35^@ cos 45^@/(sin 45^@cos 35 ...

2018年3月5日 · #x=16sin20*sin40*sin60*sin80# = #16*(sqrt3)/2*sin80*sin40*sin20# = #8sqrt3*sin80*sin40*sin20# = #4sqrt3*(cos40-cos120)*sin20#

2018年5月21日 · 24. Here, "the Nr."=96sin65^@sin35^@sin80^@, =48{2sin65^@sin35^@}sin80^@, =48{cos(65^@-35^@)-cos(65^@+35^@)}sin80^@, =48{cos30^@-cos100^@}sin80^@, =48{sqrt3/2-cos(180 ...

2018年8月10日 · The value of #sin20.sin40.sin60.sin80# is? Trigonometry. 1 Answer maganbhai P. Aug 10, 2018 #sin20 ...

2017年5月9日 · Now sin80 can be written as Sin(90-10) which is equal to cos10° Similarly sin100 can be written as Sin(90+10) which is in second quad and hence can be written as Cos10° so , sin80 = sin100 But 80 can't be equal to 100 because trignometry includes quadrants and on using it both angles comes out to be 10° and hence

2016年9月26日 · This formula can be simplified to -sin60=-sqrt(3)/2 sin20cos80-cos20sin80=sin(20-80)=sin(-60)=-sin60=-sqrt(3)/2

2017年8月8日 · Prove that Sin10°+Sin20°+Sin40°+Sin50°=Sin70°+Sin80° - 1370752. Brainly User Brainly User ...

2017年12月2日 · As #cos2A=cos^2A-sin^2A=2cos^2A-1=1-2sin^2A#. #cos^2 20^@-sin^2 20^@=cos(2xx20^@)=cos40^@=p# Also #cosA=sqrt((1+cos2A)/2)#

2017年12月25日 · Let's let #/_BDC= /_x# and #/_DBC= /_y#. Since triangle #ABC# is an isosceles triangle then #/_ACB=80^@#. Then from triangle #BDC# we have:

2017年6月11日 · Sin10+sin20+sin40+sin50=sin70+sin80 Get the answers you need, now!